Try searching the internet for Langley's Problem. It is similar-I solved it some time ago using trigonometry.
This is a tedious drawing so if you are going to procede I'll thank you in advance. Here is my reasoning for solving this problem.Draw an isosceles triangle ABC with Side AB = Side AC. Draw a line from C to side AB and label that line CD. Now draw a line from B to side AC. Label that line BE. Let angle EBC = 60 degrees, angle BCD equal 70 degrees, angle ABE equal 20 degrees, and angle DCE equal 10 degrees. Now draw line DE.
Find what angle EDC is by using geometry only and no trigonometry.
Don't cheat and use protractors/rulers/all that stuff either! Go by pure geometric reasoning
Call the intersection of BE and CD point M. Let ADE=x, EDM=y, DEA=z, and DEM=140-z.
Thus we have the following system of equations:
(x+y)+20+10=180
x+20+z=180
50+y+(140-z)=180
And when I solve this I get a large angle, a negative angle, and an angle of zero. I think my work is sound. Please look it over.
Thanks
This problem is not designed to be treated as a high-school angle solving problem. The trick it to draw a contrusction and even from there it is difficult.Originally Posted by Jameson
BTW if your system is consistent, then all solutions can be expressed though a parameter . Then all you need to to find a solution that is 'reasonable' since there are infinitely many.
The answer is 20. The question was asked on www.collegeconfidential.com on the UPenn forum, and the OP hasn't posted an explanation to the solution to this problem. Here is a link that will guide you though. http://agutie.homestead.com/files/LangleyProblem.html